Question 628051
The scores on a certain test are normally distributed with a mean score of 51 and a standard deviation of 2. What is the probability that a sample of 90 students will have a mean score of at least 51.2108
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z(51.2108) = (51.2108-51)/[2/sqrt(90)] = 1
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P(x >= 51.2108) = P(z >= 1) = 0.1587
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Cheers,
Stan H.